Answer:
maybe you can use Grammarly
Explanation:
ASSUME the availability of a variable, stdin, that references a Scanner object associated with standard input.
Answer:
The JAVA program is as follows.
import java.util.Scanner;
public class Program
{
static int n;
public static void main(String[] args) {
//scanner object
Scanner stdin = new Scanner(System.in);
//loop executes till number entered is in the range of 1 to 10
do
{
System.out.print("Enter any number: ");
n = stdin.nextInt();
}while(n<1 || n>10);
System.out.println("Number is valid. Exiting...");
}
}
OUTPUT
Enter any number: 0
Enter any number: 23
Enter any number: 10
Number is valid. Exiting...
Explanation:
The program is explained.
1. The integer variable, n, is declared static since the variable should be accessible inside main() which is static.
2. An object of Scanner, stdin, is created inside main().
3. Inside do-while loop, the user is asked input any value for n.
4. The loop continues till user inputs a number which is not in the given range, beginning from 1 to 10.
5. The loop will not be terminated till the user enters a valid input.
6. Once a valid input is entered, the message is displayed to the console and the program ends.
7. The variable, n, is declared at the class level and is a class variable. Class variables are declared inside the class but outside all the methods in that class.
8. Since JAVA is a purely object-oriented language, all the code is written inside class.
9. The object of class is not created since only one class is included in the program. If more than one class is included, then the object of the class which does not has the main() method will be created inside main().
10. User input for integer is taken via scanner object and using nextInt() method.
11. The methods, print() and println() display the messages to the console. The only difference is that println() inserts a new line after displaying the message.
12. The name of the class having main() and the name of the program should be the same.
b. Backstage
c. Web Layout
d.Both a and b
To close the backstage view, click File on the Ribbon or click the preview of the document in the Info gallery to return to the document window. Thus, the correct option for this question is B.
A document window may be defined as a kind of enclosed child window that is significantly linked with a multiple-document interface (MDI) window. These types of windows are typically utilized for the display and modification of source code or text, but they can also host other functional types.
According to the context of this question, the backstage view displays the list of the most recent files that have been worked on. And if you want to open these files you must use this view which generally saves your time.
Therefore, to close the backstage view, click File on the Ribbon or click the preview of the document in the Info gallery to return to the document window. Thus, the correct option for this question is B.
To learn more about Backstage view, refer to the link:
#SPJ2
equality
alignment
contrast
Answer:
contrast
Explanation:
Answer:
I'm pretty sure it's contrast
Explanation:
It would be contrast because when you're looking for a highlighting color, you'd want to use colors that stand out from the rest. This helps you find what you want to see again very easily.
I'm sorry if this is wrong, but if it's right pls vote me brainliest
Answer:
Program in c++ and java
Explanation:
C++ Code
#include<iostream> //for input and output
#include <string> // for string
using namespace std;
int main()
{
string firstname;
string lastname;
cout<<"Please input your first name :";
cin>> firstname;
cout<<"Please input your last name :";
cin>> lastname;
cout<< "Your name is "<< firstname << " "<<lastname;
// declaring output string stream
return 0;
}
Java Code
import java.util.Scanner;
public class SquareRoot {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
System.out.print("Please enter your first name :");
String firstName = scanner.next();
System.out.print("Please enter your last name :");
String lastname = scanner.next();
System.out.println("Your name is "+firstName+" "+lastname);
}
}
Output
Please input your first name :John
Please input your last name :Stone
Your name is John Stone
3.How do you benefit from that part or feature?